SOLUTION: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to 3x – 6y = 9 and passing through (-2, -1).

Algebra ->  Graphs -> SOLUTION: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to 3x – 6y = 9 and passing through (-2, -1).      Log On


   

Question 107806: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to 3x – 6y = 9 and passing through (-2, -1).
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
3x-6y=9
-6y=-3x+9 NOW WE HAVE TO REDUCE THE -6Y TERM TO Y.
WE DO THIS BY DIVIDING ALL THE TERMS BY -6.
-6y/-6=-3x/-6+9/-6 NOW WE REDUCE WHENEVER POSSIBLE.
y=1/2x-3/2 HERE IT'S EASIER TO DEAL IN DECIMALS THAN FRACTIONS.
y=.5x-1.5 (red line)
the perpendicular line has a slope of -2 ( THIS IS THE NEGATIVE RECIPRICAL OF .5) & PASSING THROUGH (-2,-1) USING THE LINE EQUATIO (Y=mX+b) WHERE m=SLOPE & b= THE Y INTERCEPT.
-1=-2*-2+b
-1=4+b
b=-1-4
b=-5
thus the equation is:
y=-2x-5 (green line)
+graph%28+200%2C+200%2C+-6%2C+5%2C+-6%2C+5%2C+y+=+.5x+-1.5%2C+y+=+-2x+-5%29+ (graph 200x200 pixels, x from -6 to 5, y from -6 to 5, of TWO functions y = .5x -1.5 and y = -2x -5).